Balance this equations for redox reactions in basic solution.

Fe(OH)3(s) + OCl- �¨ FeO42- + Cl-

no its not right, its a basic solution so no H+ but OH-

If you don't know the system for balancing redox equations you need to learn it. This is what I use and I will give you one of the equation.

Step 1. Writ the half equation.
OCl^- ==> Cl^-

Step 2. Write oxidations states of elements and add electrons to account for change in oxidation state.
Cl is +1 on left and -1 on right; therefore, add 2electrons to the left side like so.

2e + OCl^- ==> Cl^-

Step 3. Count charge and add OH (in this case but add H^+ for acidic solutions) to the appropriate side.

charge on left is -3; on right is -1 which means add 2OH to right side.
2e + OCl^- ==> Cl^- + 2OH^-

Step 4. Add H2O (usually to the opposite side where OH or H were added).
H2O + 2e + OCl^- ==> Cl^- + 2OH^-

Step 5. Check to make sure it is balanced. ALWAYS CHECK IT.

You do the Fe the same way. It will be easier to do, I think, if you separate the half equation like this.
Fe^+3 ==> FeO4^-2

Sure, let's balance this equation in a humorous way.

Fe(OH)3(s) + OCl- ⟶ FeO42- + Cl-

Well, Fe (iron) sounds like it could use a little pick-me-up, so let's start by adding some OCl- (hypochlorite) to its life. Fe takes 3 OCl- to form FeO42-. That should put a smile on Fe's face.

Fe(OH)3(s) + 3OCl- ⟶ FeO42- + Cl-

Now, let's make sure Cl (chlorine) feels included. We can add 6 Cl- ions on the other side, because why not give chlorine a full squad of atoms to hang out with?

Fe(OH)3(s) + 3OCl- ⟶ FeO42- + 6Cl-

But wait! We're in a basic solution, which means we have to balance out the charges. So we need to add some OH- (hydroxide) to neutralize everything.

Fe(OH)3(s) + 3OCl- + 3OH- ⟶ FeO42- + 6Cl- + 3H2O

And voila! Our equation is now balanced, and Fe and Cl are happily balanced buddies.

To balance this redox equation in basic solution, you'll need to follow these steps:

Step 1: Assign oxidation states to all the elements in the equation.
In this equation, Fe(OH)3 has an oxidation state of +3, OCl- has an oxidation state of -1, FeO42- has an oxidation state of +6, and Cl- has an oxidation state of -1.

Step 2: Separate the equation into two half-reactions, one for the oxidation half-reaction and one for the reduction half-reaction.

Oxidation half-reaction:
Fe(OH)3(s) → FeO42- + 3H2O

Reduction half-reaction:
OCl- + 2H2O → Cl- + 4OH-

Step 3: Balance all elements except for hydrogen and oxygen in each half-reaction. Start by balancing the atoms other than hydrogen and oxygen.

Oxidation half-reaction:
Fe(OH)3(s) → FeO42- + 3H2O
Balanced so far.

Reduction half-reaction:
OCl- + 2H2O → Cl- + 4OH-
Add 2 OH- ions to the product side:
OCl- + 2H2O → Cl- + 4OH-

Step 4: Balance oxygen atoms by adding water (H2O) molecules to the side that needs oxygen.

Oxidation half-reaction:
Fe(OH)3(s) → FeO42- + 3H2O
Balanced.

Reduction half-reaction:
OCl- + 2H2O → Cl- + 4OH-
Add 2 H2O molecules to the reactant side:
OCl- + 2H2O → Cl- + 4OH- + 2H2O

Step 5: Balance hydrogen atoms by adding hydrogen ions (H+). The number of H+ ions needed is equal to the number of H2O molecules added in Step 4.

Oxidation half-reaction:
Fe(OH)3(s) → FeO42- + 3H2O
Balanced.

Reduction half-reaction:
OCl- + 2H2O → Cl- + 4OH- + 2H2O
Add 8H+ ions to the reactant side:
OCl- + 2H2O + 8H+ → Cl- + 4OH- + 2H2O

Step 6: Balance the charges in each half-reaction by adding electrons (e-). The total charge on each side of the equation should be the same.

Oxidation half-reaction:
Fe(OH)3(s) → FeO42- + 3H2O
Balanced.

Reduction half-reaction:
OCl- + 2H2O + 8H+ + 7e- → Cl- + 4OH- + 2H2O
Add 7e- to the reactant side:
OCl- + 2H2O + 8H+ + 7e- → Cl- + 4OH- + 2H2O

Step 7: Multiply both half-reactions by appropriate coefficients to equalize the number of electrons transferred in each half-reaction.

Multiply the oxidation half-reaction by 7:
7Fe(OH)3(s) → 7FeO42- + 21H2O

Multiply the reduction half-reaction by 1:
OCl- + 2H2O + 8H+ + 7e- → Cl- + 4OH- + 2H2O

Step 8: Add the two half-reactions together and cancel out any common species on both sides of the equation.

7Fe(OH)3(s) + OCl- + 2H2O + 8H+ + 7e- → 7FeO42- + Cl- + 4OH- + 2H2O

Simplify the equation:
7Fe(OH)3(s) + OCl- + 8H+ + 7e- → 7FeO42- + Cl- + 4OH- + 2H2O

Step 9: Balance the electrons transferred in the overall reaction by multiplying one of the half-reactions by an appropriate coefficient to equalize the number of electrons.

Multiply the oxidation half-reaction by 7:
7Fe(OH)3(s) + OCl- + 8H+ + 7e- → 7FeO42- + Cl- + 4OH- + 2H2O

The balanced equation for the redox reaction in basic solution is:
7Fe(OH)3(s) + OCl- + 8H+ → 7FeO42- + Cl- + 4OH- + 2H2O

I got the following, but I'm not sure if I did it correctly:

2Fe(OH)3 + 3OCl- ->2FeO4-2 + 3Cl- + 4H+ + H2O

If someone could let me know whether I solved it correctly or not, that'd be nice. :)